How to find the inverse operator?

Question

Let $A: C([0,1])\to C([0,1])$ be a linear operator defined with: $A(x(t)) = x(t) + \int_0^t x(s)\,ds$.

It is actually easy to see that $A$ injective is, but it a bit of problem to show that it's bijective. Though, the main puzzle is: how to find the inverse operator? I myself tried but didn't far away of the expression

$$ A^{-1} (x(t)) = A^{-1} \int_0^t x(s)\,ds - x(t) $$

Thanks in forward!

Answer 1

If $f=Ax$ and we assume for the moment that $x$ is differentiable, from $$ f(t)=x(t)+\int_0^t x(s)\,dx, $$ we get $$ f'=x'+x. $$ This is a linear differentiable equation on $x$, with solution $$ x(t)=x(0)e^{-t}+e^{-t}\int_0^te^sf'(s)\,ds. $$ Of course this only works if $f$ is differentiable. But, using parts, $$ \int_0^te^s f'(s)\,dx=e^tf(t)-f(0)-\int_0^te^sf(s)\,ds, $$ so we may write $$\tag1 x(t)=x(0)e^{-t}+f(t)-e^{-t}f(0)-e^{-t}\int_0^te^sf(s)\,ds $$ Now, $Ax(0)=x(0)$, so $x(0)=f(0)$ above. So $(1)$ says that $$\tag2 \boxed{(A^{-1}y)(t)=y(t)-e^{-t}\int_0^te^sy(s)\,ds. } $$ It is now an easy exercise to check that $(2)$ is actually the inverse of $A$, for all $x\in C[0,1]$.

Answer 2

Consider $A$ to be a perturbation of the identity operator: $A = I - B$, where $$ B(x)(t) := -\int\limits_{0}^{t} x(s) \, ds. $$ If we can prove that the Neumann series $$ \tag{$*$} \sum\limits_{k=0}^{\infty} B^k $$ converges in the operator norm, we are done: then $$ A^{-1} = (I - B)^{-1} = \sum\limits_{k=0}^{\infty} B^k. $$ We have the following equality $$ \lVert B^k \rVert = \frac{1}{k!} $$ (for a proof, see, e.g., Norm of integral operator), hence the Neumann series $(*)$ is convergent.